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PhysicsInteger+4Question 20 of 60

IIT-JEE 2012 · Paper 1, Physics question 20

A proton is fired from very far away towards a nucleus with charge Q=120eQ=120e, where ee is the electronic charge. It makes a closest approach of 10 fm to the nucleus. The de Broglie wavelength (in units of fm) of the proton at its start is: (take the proton mass, mp=(5/3)×10−27m_p=(5/3)\times10^{-27} kg; h/e=4.2×10−15h/e=4.2\times10^{-15} J.s/C; 14πε0=9×109\dfrac1{4\pi\varepsilon_0}=9\times10^9 m/F; 1 fm =10−15=10^{-15} m)

Answer and worked solution

Official answer: 7

Initial kinetic energy equals the potential energy at closest approach (head-on):
K=(120e)e4πε0r.K=\frac{(120e)e}{4\pi\varepsilon_0r}.

Momentum. p=2mpKp=\sqrt{2m_pK}, so
λ=hp=he4πε0r2mp⋅120=4.2×10−1510−149×109×2×53×10−27×120.\lambda=\frac hp=\frac h{e}\sqrt{\frac{4\pi\varepsilon_0r}{2m_p\cdot120}}=4.2\times10^{-15}\sqrt{\frac{10^{-14}}{9\times10^9\times2\times\frac53\times10^{-27}\times120}}.
The square root is 10−143.6×10−15=2.78=1.667\sqrt{\dfrac{10^{-14}}{3.6\times10^{-15}}}=\sqrt{2.78}=1.667.

Result. λ=4.2×10−15×1.667=7.0×10−15\lambda=4.2\times10^{-15}\times1.667=7.0\times10^{-15} m =7=7 fm.

Answer: 7

Topics:de Broglie wavesPotential energy of charge systems
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